Littlewood polynomials
Bachelor’s thesis
July 2015
Student: Majken Roelfszema s2350629 Primary supervisor: prof. dr. J. Top
Abstract
A Littlewood polynomial is a polynomial of which all coefficients are in {−1, 1}. The zeros of Littlewood polynomials do not only make a beautiful plot, they also have some fascinating properties. For example the symmetry in the real and imaginary axis and the unit circle and bounds for the absolute value of the zeros. Also, the intersection of the set of zeros of Littlewood polynomials with the set ∪
d∈QQ(
√
d) turns out to consist of only 12 elements.
We sketch the proofs of two known facts, namely that the closure of the set of zeros of Littlewood polynomials is connected and that all z with 2−14 ≤ |z| ≤ 214 are in that closure.
Contents
1 Introduction 3
2 Properties and proofs 5
3 Square roots 11
4 The closure is connected 13
5 A ring in the closure 17
6 Properties to be studied 22
7 Conclusion 24
1 Introduction
In the Dutch mathematics journal "Pythagoras" [1] of Februari 2015, there was an article about the zeros of Littlewood polynomials. Andrej Bauer made a plot of the zeros of Littlewood polynomials with degree up to 26. This plot was very beautiful because of its symmetries.
A Littlewood polynomial is a polynomial with all coefficients in the set {−1, 1}. A Littlewood polynomial can be written as follows:
`(z) =
n
X
j=0
ajzj = a0+ a1z + . . . + anzn= ±1 ± z ± . . . ± zn.
In figure 1 the zeros of Littlewood polynomials with degree n ≤ 30 are plotted in the complex plane.
Figure 1: The zeros of Littlewood polynomials with degree n ≤ 30 The picture indicates that the set of zeros is symmetrical in the x-axis, the y-axis and in the unit circle. This will be proven in section 2, along with some other properties that are not too difficult to prove.
John Edensor Littlewood was a British mathematician who worked in the field of mathematical analysis. It was around 1950 that Littlewood examined polynomials with coefficients from the set {−1, 1}, according to [2] . This is the reason those polynomials are nowadays called Littlewood polynomials.
In recent studies, the zeros of some special Littlewood polynomials are also used, but under the name of Multinacci numbers. A Multinacci number tm is defined as the unique real root of the equation
xm+ xm−1+ . . . + x = 1 which lies in 12, 1. [3] Also, the number t1
m is called a Pisot-Vijayaraghavan number.
The zeros of Littlewood polynomials have a lot of fascinating properties.
The aim of this thesis is to get an overview of some properties of the zeros
of Littlewood polynomials. The main result, first proven by T. Bousch in 1988 [4], is that the closure of the set of zeros of Littlewood polynomials is connected.
Throughout this thesis, the set of zeros of Littlewood polynomials will be called D. In section 2, as mentioned, some properties are stated and proven.
In section 3, the set ∪
d∈QQ(
√d) will be examined. Two known results, that the
closure of D is connected and that the annulus n
2−14 ≤ |z| ≤ 214o
is entirely contained in the closure of D, will be shown in section 4 and 5 respectively.
Some related concepts and pictures are considered in section 6.
2 Properties and proofs
In this section some properties of Littlewood polynomials will be stated and proven. The set of zeros of Littlewood polynomials is denoted by D.
Theorem 1. All the zeros of Littlewood polynomials are in the annulus {12 < |z| < 2}.
In order to prove Theorem 1, we will prove two lemma’s.
Lemma 2. The zeros of Littlewood polynomials satisfy |z| < 2.
Proof of Lemma 2: Assume α is a zero of a Littlewood polynomial of degree n. We have:
| ± αn± αn−1± . . . ± α ± 1| = 0
If |α| ≤ 1, we have |α| < 2, so we can assume |α| > 1. This gives:
| ± αn± αn−1± . . . ± α ± 1|
≥ |α|n−1 + |α| + . . . + |α|n−1
= |α|n (
1 −
n
X
i=1
1
|α|i )
> |α|n (
1 −
∞
X
i=1
1
|α|i )
= |α|n (
1 + 1 − 1 1 −|α|1
)
= (∗)
The previous term was obtained because the summation mentioned is:
∞
X
i=1
1
|α|i = (
∞
X
i=0
1
|α|i) − 1 = 1
1 −|α|1 − 1, for 1
|α| < 1 i.e. |α| > 1.
Thus it follows that:
(∗) = |α|n
2 |α| − 1
|α| − 1
− |α|
|α − 1|
= |α| − 2
|α| − 1
|α|n (1)
This implies that equation 1 is smaller than zero and from this it follows that
|α| < 2. (Equation 1 is defined because |α| < 1). This proves Lemma 2. The theorem will be proven with the following lemma.
Lemma 3. If α is a zero of a Littlewood polynomial, so is α−1.
Proof of Lemma 3: Let α be a root of a Littlewood polynomial `(z) = Pn
j=0ajzj. Put `r(z) = zn· `(z−1) =Pn
k=1akzn−k, which is also a Littlewood polynomial. Note that α 6= 0, because the constant term in ` is nonzero. It follows that
`r(α−1) = α−n· `(α) = 0.
So α−1 is also the zero of a Littlewood polynomial, α−1 ∈ D.
If β ∈ D, with β ≤ 12, then β−1 ∈ D. But it holds that |β−1| ≥ 2, so according to lemma 2, β cannot be a zero of a Littlewood polynomial.
Therefore every zero of a Littlewood polynomial satisfies |z| > 12.
Finally it can be concluded that D ⊂ {12 < |z| < 2}, which proves Theorem 1.
Theorem 4. D is symmetric in the real axis, in the imaginary axis and in the unit circle.
For this theorem to hold, it should hold that the following three bijections send elements of D to elements of D.
α : z → z β : z → −z
γ : z → 1 z
Proof for the map α: Let α be a zero of a Littlewood polynomial of degree n. Let α be its complex conjugate.
`(α) = 0 →
n
X
j=1
ajαj = 0.
We show that α is a zero of the same polynomial:
`(α) =
n
X
j=0
ajαj =
n
X
j=0
ajαj =
n
X
j=0
ajαj =
n
X
j=0
ajαj = 0 = 0.
It follows that `(α) = 0.
Proof for the map β: Let α be a zero of a Littlewood polynomial of degree n: l(α) = 0. Then there exists a Littlewood polynomial for which −α is a zero. Write
`(z) =
n
X
j=1
ajzj, and put
`−(z) =
n
X
j=1
aj(−z)j.
The polynomial `− therefore has the same coefficients as ` for even j and −aj for odd j. This polynomial `− is also a Littlewood polynomial of degree n.
It holds that `−(−α) = 0 whenever `(α) = 0. So for every zero of a Littlewood polynomial α, it holds that −α is also a zero of a Littlewood polynomial (and the polynomials have the same degree).
Proof for the map γ: Let α be a zero of a Littlewood polynomial of degree n. The reciprocal of
`(z) =
n
X
j=0
ajzj is defined as
`r(z) = zn`(1 z).
The reciprocal `r is a Littlewood polynomial which was already used in the proof of Lemma 3. If α is a zero of `(z), 1α is a zero of `r(z):
αn` 1
1 α
!
= αn`(α) = 0.
These three maps together generate a group of eight elements {id, α, β, γ, αβ, αγ, βγ, αβγ}
with the following properties:
α2 = β2 = γ2 = id (αβ)2= (αγ)2 = (βγ)2 = id αβ = βα, αγ = γα, βγ = γβ.
The three maps α, β and γ generate a group that is isomorphic to Z/2Z × Z/2Z × Z/2Z.
According to [5], the inverse of a complex number z in a circle with radius a is
a2 z .
It follows that D is symmetric in the unit circle if the following holds:
z ∈ D ⇔ 1 z ∈ D.
It is now clear that this is true and Theorem 4 follows.
Theorem 5. D ∩ Q = {−1, 1}.
Theorem 5 states that the only real rational zeros of Littlewood polyno- mials are −1 and 1. In section 3 we will expand on this, by taking Q(√
d) for d ∈ Q. Theorem 5 can be proven using the Lemma of Gauss.
Lemma 6 (Lemma of Gauss [6]). If a non-constant polynomial with integer coefficients is irreducible over the integers, then it is also irreducible if con- sidered as a polynomial over the rationals.
It follows from the Lemma of Gauss that: if f = xn+an−1xn−1+· · · ∈ Z[X]
is a monic polynomial and h ∈ Q[X] is also a monic polynomial with h|f , then h ∈ Z[x].
From this it follows that if r ∈ Q is a zero of the polynomial g = xn+ an−1xn−1+ . . . + a0 ∈ Z[x], then r ∈ Z. This is because if r ∈ Q is a zero of g, then the monic polynomial x − r divides g. This proves Theorem 5.
Theorem 7. Every e2πir with r ∈ Q is an element of D.
This is an immediate consequence of the much stronger result which is stated as follows.
Theorem 8. For every e2πir with r ∈ Q and every m ∈ N there exists a Littlewood polynomial with ordz=e2πir`(z) ≥ m.
Proof of Theorem 8: Let e2πiab with a, b ∈ Z be an element of D.
Let `(z) be a Littlewood polynomial of degree n. The equation zn+1`(z) has coefficients in {−1, 1} for zn+1, zn+2, . . . , z2n+1 and zj = 0 for all other values of j. It follows that
±zn+1`(z) ± `(z) = ±(zn+1± 1)`(z)
is again a Littlewood polynomial. In the same way, this holds for the polyno- mials
`b(z) = (zb−1+ . . . + 1)(zb− 1)(z2b− 1)(z4b− 1)(z8b− 1) · . . . , with the property
`b(e2πiab) = 0.
It is known that e2πic with c ∈ Z gives 1. We have zb = 1, z2b = 1, z4b = 1, . . . and so the polynomial `b has a zero in e2πiab of some order ord
z=e2πi ab`b(z) which can be controlled.
The degree of `b is n = b − 1 + b + 2b + 4b + . . ..
A Littlewood polynomial with ordz=e2πir`(z) ≥ m, r ∈ Q written as r = ab, can be constructed as follows:
`m(z) = (zb−1+ . . . + 1)(zb− 1)(z2b− 1) · . . . · (z2m−1b− 1).
This polynomial `m(z) has ordz=e2πir`m(z) ≥ m. This proves Theorem 8 and therefore also Theorem 7. Because Theorem 7 holds, the unit circle is contained in the closure of D, D. In section 5 we will see that in fact an annulus,
n
2−14 ≤ |z| ≤ 214o
, is contained in D.
Theorem 9. 1 ∈ D is a limit point of D.
In figure 2 up to 4 we see that for higher degree, the roots of Littlewood polynomials are also closer to 1.
For 1 to be a limit point of D means that ∀ > 0, ∃ an element z 6= 1 of D with z ∈ B(1). Take α ∈ D, α > 0. For the proof of Theorem 9, the following lemma can be used.
Figure 2: Degree 30 Figure 3: Degree 40 Figure 4: Degree 50
Lemma 10. α ∈ D ⇔ √n
α ∈ D ∀n ∈ N, for every choice of √n α ∈ C.
(⇒): Let `(z) be a Littlewood polynomial with `(α) = 0, let n ∈ N and let √n
α be an nth power root of α. Then
`n(z) = (1 + z + z2+ . . . + zn−1) · `(zn) is also a Littlewood polynomial and `n(√n
α) = 0.
(⇐): The above is also true for n = 1. This proves Lemma 10.
For every > 0, an n ∈ N can be found such that √n
α ∈ B(1). Therefore 1 is a limit point of D.
Theorem 11. 2 is a limit point of D.
To prove this, it is sufficient to show that ∀ there is an element of D in B(2). Where B(2) is an open ball of radius with 2 as the center. Let
`(z) = zn− zn−1− zn−2− . . . − 1.
This polynomial has `(2) = 1. Because polynomials are continuous func- tions, if there would be a number very close to 2 for which ` is negative, there would be a zero inbetween 2 and this number. The Littlewood polynomial ` can be written as follows.
`(z) = zn−1 − xn
1 − x = xn+1− xn+ 1 − xn
x − 1 = xn(x − 2) + 1 x − 1 Assume 0 < << 1.
`(2 − ) = (2 − )n· (−) + 1 1 − Let = 2n and take n sufficiently large. Then
(2 − )n=
2 − 2
n
n
= 2n
1 − 1
n
n
. Since log 1 − 1nn
= n log 1 −n1 converges to 1 for n → ∞, we have 2n
1 − 1
n
> 2n· c
for some constant c > 0 and hence
`
2 − 2
n
< −2n+1n·c + 1
1 −n2 < 0, ∀n >> 0.
It follows that `(z) has a zero in between 2−n2 and 2, so in B2
n(2). Therefore 2 is a limit point of the set D and 2 ∈ D. It also follows that 12 is a limit point of the set D, because of the symmetry (Theorem 4).
Theorem 12. If r ∈ Q≥0, then √
r ∈ D ⇔ r = 1.
(⇐): This is trivial, because ±1 ∈ D.
(⇒): Assume r ∈ Q≥0. If r has √
r ∈ Q, it follows from Theorem 5.
Otherwise, the minimal polynomial for √
r over Q is x2 − r and Lemma 6 implies that r ∈ Z.
Because √
r ∈ D, we have:
±√ rn±√
rn−1± . . . ±√
r = 1
√r
±√
rn−1± . . . ± 1
= 1.
Both√
r and (±√
rn−1± . . . ± 1) are elements of Z[√ r],√
r ∈ Z[√
r]×. Let N denote the norm. It follows that
N (√
r) · N (±√
rn−1± . . . ± 1) = 1.
Since these norms are natural numbers, N (√
r) = r = 1.
This proves Theorem 12.
This Theorem is in fact one of the cases considered in the next section.
There we do not only consider the square roots of a rational number, but the set Q(√
d) for d ∈ Q.
3 Square roots
In this section the intersection of the sets Q(√
d) and D will be considered, where d ∈ Q. Moreover, ∪
d∈QQ(
√
d) will be considered. First of all, we can assume that d ∈ Z, because d can just be multiplied by an integer until it is.
Also, d should not be a square or divisable by a square > 1, because if√ d ∈ Z, Q(
√
d) = Q.
This leaves us with the following choices for d:
d ∈ {. . . , −5, −3, −2, −1, 2, 3, 5, 6, 7, 10, 11, 13, 14, . . .} . If a + b√
d ∈ Q(√
d) ⊂ C is a zero of a Littlewood polynomial, then the minimal polynomial of a+b√
d is in Z[x] (this follows from the Lemma of Gauss [6]) and has degree 1 or degree 2. If it has degree 1, b = 0 and a + b√
d ∈ Q, which according to Theorem 5 in section 2 means that a + b√
d ∈ {−1, 1}.
If however the degree of the minimal polynomial is 2, b 6= 0 and the minimal polynomial is
x2− 2ax + a2− db2, ∈ Z[x], so 2a ∈ Z and a2− db2 ∈ Z. Then a ∈ 12Z and b ∈ 12Z.
Also, a + b√
d has an inverse which also has a minimal polynomial in Z[x].
Let this inverse be n + m√
d. Then also n, m ∈ 12Z. Let N denote the norm.
It follows that
N (a + b
√
d) = a2− db2 ∈ Z N (n + m√
d) = n2− dm2∈ Z and because n+m√
d is the inverse of a+b√
d, N (a+b√
d)·N (n+m√ d) = 1, so a2− db2 = ±1.
Because a + b√
d ∈ D and a, b ∈ 12Z, also a − b
√
d ∈ D, −a − b√
d ∈ D and
−a + b√
d ∈ D (Theorem 4, section 2). In other words: ±a ± b√ d ∈ D.
It also follows from the fact that a + b√
d is in D that if we let x = a + b√ d,
±xn± xn−1± . . . ± x = ∓1 x(±xn−1± . . . ± 1) = ∓1 where x ∈ Z[√
d] and (±xn−1± . . . ± 1) ∈ Z[√
d], so x = a + b√
d ∈ Z[√ d]×. If d < 0, three distinct cases occur.
For d = −1, the group Z[√
−1]× is {1, −1, i, −i}. Because these are all in the set D,
Q(
√−1) ∩ D = {1, −1, i, −i}.
For d = −3, d ≡ 1 mod 4. We take:
1, −1,1 2 +1
2
√−3, −1 2 +1
2
√−3, −1 2 −1
2
√−3,1 2−1
2
√−3
. These are all in D, and for no other values of a and b is a + b√
−3 ∈ D, so Q(
√−3) ∩ D =
1, −1,1 2 +1
2
√−3, −1 2 +1
2
√−3, −1 2 −1
2
√−3,1 2−1
2
√−3
.
For all other negative values of d, the intersection is Q(√
d) ∩ D = {−1, 1}.
(The groups Z[√
d]× are equal to {−1, 1}.)
Moving on to positive values of d. Choose the largest of ±a ± b√ d ∈ D:
|a| + |b|√
d. It holds that
|a| + |b|√ d ≥ 1
2 +1 2
√ d.
Because this is ≥ 2 for all d ≥ 9, we have that Q(√
d) ∩ D = {−1, 1} for all d ≥ 9.
For the remaining values of d > 1, there are two cases: the case where d ≡ 1 mod 4 and the case where d ≡ 2 mod 4 or d ≡ 3 mod 4.
Let d ≡ 2 mod 4 or d ≡ 3 mod 4. If a + b√
d ∈ Z[√
d]×, then {a + b√
d, −a + b
√
d, −a − b
√ d, a − b
√
d} ∈ Z[
√ d]×
and these units are distributed along the real axis as in the figure below.
The group Z[√
d]×is equal to (−1, ) where is the smallest unit with > 1.
For the case where d = 2, we get: Z[√
2]× = {. . . , −1 −√
2, −1, 1, 1 +√ 2, . . .}
and because 1 +√
2 > 2, D ∩ Q(√
2) = {−1, 1}.
For all other values of d with d ≡ 2 mod 4 or d ≡ 3 mod 4, the same follows: D ∩ Q(√
d) = {−1, 1}.
Let d ≡ 1 mod 4. If for a + b√
d, a, b ∈ Z, the same as above holds. If, on the other hand, a and b are both not in Z, we get the |a| + |b|√
d with a, b ∈ 12Z.
For d = 5, this gives that Q(√
5)∩D =n
−1+
√ 5
2 , −1,1−
√ 5 2 , −1−
√ 5 2 , 1,1+
√ 5 2
o . This concludes the final case. The following theorem can be concluded.
Theorem 13. D ∩
∪
d∈QQ(
√d)
=±1, ±i, ±12 ±12√
−3, ±12 ±12√ 5 .
4 The closure is connected
In this chapter, a proof of Theorem 14 is given. This argument is due to T.
Bousch (1988)[4]. Some parts are more involved and will only be sketched.
Theorem 14. D is connected.
An important lemma that will be used in the proof and in the next section is Lemma 15. Firstly, three sets will be defined. They are subsets of the set O(B1(0)), the space of analytic functions on the open disc |z| < 1, equipped with ’the topology of compact convergence’.
Compact convergence: a sequence or series converges compactly in X if it converges uniformly on every compact subset of X.
W =
f (z) =
∞
P
i=0
bizi, b0= 1, ∀i, |bi| ≤ 2
H =
f (z) =
∞
P
i=0
bizi, b0 = 1, ∀i, bi ∈ {−1, 0, 1}
H1 =
f (z) =
∞
P
i=0
bizi, b0= 1, ∀i, bi ∈ {−1, 1}
Lemma 15. Let y ∈ (0, 1) and > 0 such that y + < 1. Then ∃n0 ∈ N such that: ∀f ∈ W , ∀s with |s| < y and f (s) = 0, then ∀g ∈ W with f − g ∈ zn0C[[z]], ∃s0 ∈ B(s) with g(s0) = 0.
Note that Lemma 15 is defined for the set W , for the proof it is necessary that it holds for H1 ⊂ W . The proof of Lemma 15 is a bit involved and is given in [4]. A sketch of the proof is given here. Firstly, the set
F =(f, s) ∈ W × By(0)|f (s) = 0
is defined. The set By(0) is the disc with |z| ≤ y. Because W × By(0) is compact, the closed subset F is also compact. (This is a Theorem from analysis.)
Secondly, the distance |f (s) − g(s)| for f, g with f − g ∈ zn0C[[z]] is given an upper bound for |s| < y + . Let f (s) =
∞
P
n=0
ansn and g(s) =
∞
P
n=0
bnsn. Then
|f (s) − g(s)| = |sn0
∞
P
k=0
(an0+k− bn0+k)sk|
≤ |s|n0 · 4 ·
∞
P
k=0
|s|k
≤ 4 · (y + )n0·
∞
P
k=0
(y + )k
= 4·(y+)1−y−n0 := ξ(y, , n0).
For (f, s) ∈ F , a circle S(s, δ) around s is defined with radius δ ≤ 2. If we then take an n0∈ Z such that ξ(y, , n0) < ν2 with ν = min
z∈S(s,δ)
|f (z)|, then (f, s) is in the set:
V =
(f0, s0) ∈ F | max
|z|≤y+|f0(z) − f (z)| < ν
2, |s0− s| < 2
.
This V has the property: ∀(f0, s0) ∈ V , ∀g ∈ W with f0 and g the same n0− 1 first coefficients, ∃s00 ∈ D(s0, ) with g(s00) = 0. So for all z ∈ S(s, δ), we have |f (z) − f0(z)| < ν2 and |f0(z) − g(z)| ≤ ξ(y, , n0) < ν2. It follows that
|g(z) − f (z)| < ν. And so |g(z) − f (z)| < |f (z)| in B
2(s).
It follows from Rouché’s Theorem that f (z) and (g(z)−f (z))+f (z) = g(z) have the same amount of roots in B
2(s), and so there is at least one s00∈ B
2(s).
Therefore |s0− s00| < , which proves Lemma 15.
Note: in [4], a proof is given for the fact that D is connected. The proof given here is another one, following the proof in [7].
We will now take a look at Littlewood series. By definition, a Littlewood series is a power series of which all coefficients are in {−1, 1}. The radius of convergence of a Littlewood series is 1, which means that for |z| < 1, the series converges.
Lemma 16. For z ∈ C, |z| < 1, z lies in D ⇔ ∃ a Littlewood series which vanishes at z.
The "if" part (⇐) is proven first.
If `(z) = a0+ a1z + . . . + adzdis a Littlewood polynomial, then there exists a Littlewood series which looks as follows.
s`(z) = `(z)
1 − zd+1 = a0+ . . . + adzd+ a0zd+1+ . . . + adz2d+1+ a0z2d+2+ . . . (2) The coefficients of a Littlewood polynomial are repeated infinitely many times in the Littlewood series. Assume the Littlewood series s` vanishes at z, with
|z| < 1. Then 1 − zd+16= 0 because |z| < 1 and so `(z) = 0.
On the other hand, take a Littlewood series which does not have a repeating sequence of coefficients and z is a zero of the series. It follows from Lemma 15 that for every , an element α of D can be found such that |z − α| < .
Therefore z ∈D.
The "only if" part (⇒) can be proven in a similar way. For z ∈ D, |z| < 1 we already saw that a Littlewood series can be constructed that vanishes at z. If z ∈ D, but z /∈ D, there is a ’sequence of polynomials’ with a zero near z that converges to a certain Littlewood series which has this root z. It is a consequence of Lemma 15. This proves Lemma 16.
The set of roots of Littlewood series is denoted as R. It holds that D∩(|z| <
1) ⊆ R. This means that every zero z with |z| < 1 of a Littlewood polynomial is also a zero of a Littlewood series, as was shown in the proof of Lemma 16.
The set of Littlewood series is denoted by L ∼= {−1, 1}N. The set L is an infinite series of elements from {−1, 1} combined with zn with n ∈ N.
Therefore it is isomorphic to {−1, 1}N. The following Lemma is a result which is not part of the proof.
Lemma 17. ∃ a bijection between L and the Cantor set.
Proof of Lemma 17: The Cantor set is constructed in the following way:
the interval [0, 1] is taken. The interval is divided into three intervals of the same length. The middle interval is deleted. In the next step, the same is done for the two remaining intervals. Then the same is done in the four intervals obtained, and so on. The construction is made clear in figure 5.
Figure 5: The Cantor set
In every step of the construction of the Cantor set, from each interval two new intervals are created. If we would call the interval with the smallest numbers −1 and the other interval 1, we have a map from the Cantor set to {−1, 1}N. Take φ to be the map that sends a sequence of {−1, 1} to the number in the Cantor set obtained by choosing the intervals as mentioned: −1 means you take the lower interval and 1 means you take the higher interval.
Following the sequence a number in the Cantor set is obtained. Also, given a number of the Cantor set, there is a unique "path" followed, so a unique sequence of {−1, 1}N. It follows that the map φ is bijective. This proves Lemma 17.
Moreover, in [7] it is claimed that with the product topology, L ∼= {−1, 1}N is homeomorphic to the Cantor set.
Let 0 < r < 1. The space M is the space of finite multisets of points with
|z| ≤ r, modulo the equivalence relation S ∼ S ∪ {p} with |p| = r. This means that every set in M is a finite set with elements that are allowed to be there multiple times.
Lemma 18. Any Littlewood series has at most finitely many roots in the disc {|z| ≤ r}. The map ρ : L → M is continuous.
Any Littlewood series is an analytical function on an open disc around 0 with a radius r smaller than its radius of convergence (which is 1). Therefore it has only finitely many roots in that disc.[8]
The map ρ sends a Littlewood series to its multiset of roots. The set L is compact, therefore ρ(L) is closed and it is a strict subset of M , this means that te set of roots of Littlewood series, R, is closed. (This is a theorem from analysis.)
It follows from Lemma 15 that D ∩ (|z| < r) is dense in R ∩ (|z| < r). It follows that D ∩ (|z| < r) = R ∩ (|z| < r).
Lemma 19. The set R ∩ {|z| < r} with r ∈ (2−14, 1) is connected.
Suppose that U ⊆ R ∩ {|z| < r} is closed and open with respect to the topology induced from the complex topology.
For a set to be connected, it must not be contained in the disjoint union of sets. The set R ∩ {|z| ≥ r} is open because R ∩ {|z| < r} is finite, so closed. If there does not exist an U ⊆ R ∩ {|z| < r} which is closed and open, R is not contained in the disjoint union of sets, so it is connected. Therefore we want to show that this set U is empty.
Define LUto be the set of Littlewood series with a root in U . Then LU ⊆ L is closed and open. Choose a Littlewood series with d such that the first d coefficients get repeated:
f (z) = a0+ a1z + . . . + ad−1zd−1+ adzd+ a0zd+1+ a1zd+2+ . . . . Suppose f ∈ LU:
f (z) = a0+ a1z + . . . + ad−1zd−1+ adzd+ . . . g(z) = a0+ a1z + . . . + ad−1zd−1+ bdzd+ . . . Then g ∈ LU too:
Let a0 = 1, then f (z) = 1 + a1z + . . . + ad−1zd−1+ adzd+ . . ., and there are two cases: ad= 1 or ad= −1. Choose:
f (z) = 1 + a1z + . . . + ad−1zd−1+ zd+ . . . . Assume that:
g(z) = 1 + a1z + . . . + ad−1zd−1− zd+ bd+1zd+1+ . . . .
˜
g(z) = (1 + a1z + . . . + ad−1zd−1)/(1 + zd) If ˜g has a root in U , so does g.
f (z) = (1 + a˜ 1z + . . . + ad−1zd−1)/(1 − zd)
˜
g(z) = 1 − zd 1 + zd
f (z)˜
Therefore, if ˜f has a root in U , then so does ˜g. In this way it follows that if one element of R ∩ {|z| < r} is in U, all of them are. This is proven in [7]. This leads to a contradiction, so the set U is empty and R ∩ {|z| < r}
is connected. Theorem 19 is proven. Thus D ∩ {|z| < r} is connected and because of the symmetry, Theorem 4 in section 2 and Theorem 20 in section 4, D is connected, which proves theorem 14.
5 A ring in the closure
In this section the following theorem will be proven. This theorem was proven by Thierry Bousch in 1988. [4]
Theorem 20. n
2−14 ≤ |z| ≤ 214o
⊂ D
In order to prove Theorem 20, two propositions and a lemma will be proven.
The space Com(C) = {K|K ⊂ C, K is compact } is a metric space with the following metric:
d(K1, K2) = maxn
inf (K2 ⊂ {z ∈ C|∃k ∈ K1, |k − z| ≤ }) , inf (K1⊂ {z ∈ C|∃k ∈ K2, |k − z| ≤ })
o .
With this metric, the minimum of the number by which one of the sets should be extended to overlap the other set is taken. Then the distance between the sets is the maximum of these two numbers. This is done be- cause if one set lies within the other, one of the minima is zero. The space Com(C) is a complete metric space.
If a ‘cauchy sequence’ in Com(C) would be defined, the limit would be where two sets have d(Kn, Kn+1) = 0, which is exactly where the sets are equal. This will be a compact set, so the limit is in Com(C), and so it is complete.
Choose s ∈ C with |s| < 1 and K = {−1, 1}. The map given by
φs,K : Com(C) → Com(C)
A ∈ Com(C) → sA + {−1, 1} = sA + 1 ∪ sA − 1
is a contracting map. This means that d(A1, A2) > d(φs,K(A1), φs,K(A2)).
It follows from Banach’s fixed point theorem [9] that there exists a unique set A with φs,K(A) = A, so A = sA + 1 ∪ sA − 1. For this set A, the following holds.
A = sA + K
= K + sA = K + s(K + sA)
= K + sK + · · · = (1 + sA) ∪ (−1 + sA)
= A+∪ A−
It follows that A(s) = A is the set of all values obtained by evaluating se- quencesP
n≥0anzn, with an∈ K = {−1, 1} in s. Some examples of connected sets A(s) are plotted in figures 6, 7 and 8. These sets have a fractal structure.
Proposition 21. If 12√
2 < |s| < 1, A(s) is connected.
Proof of Proposition 21: Choose > 0, and define A to be the extension by of A: so A = {b ∈ C|∃a ∈ A, |a − b| ≤ }. This set A is mapped to itself with the maps
s1 : z → sz + 1 s2 : z → sz − 1
and A is compact. With the Lebesque measure, denoted by µ here, the following holds.
Figure 6: A(0.5 + 0.5i) Figure 7: A(0.6 + 0.3i) Figure 8: A(0.7 + 0.2i)
µ(A) ≥ π2 > 0 µ(s1(A)) = µ(s2(A))
= |s|2· µ(A)
µ(s1(A)) + µ(s2(A)) = 2 · |s|2· µ(A)
> µ(A) because |s|2> 12
⇒ s1(A) ∩ s2(A) 6= ∅
The implication above is true because otherwise µ(s1(A)) + µ(s2(A)) would be equal to µ(A). Because the space Com(C) is complete, we can take the limit of going to zero and it follows that
s1(A) ∩ s2(A) 6= ∅ and so A+∩ A−6= ∅
In [4], Lemma 1 it is stated that: A is connected ⇔ A+∩ A− 6= ∅. It is also proven there. It follows that A is connected and so for 12√
2 < |s| < 1, A(s) is connected. So Proposition 21 follows.
Proposition 22. If A(s2) is connected, then A(s) contains the element 0.
The sets M and M1 are defined as follows:
M = {s ∈ B1(0)|A(s) is connected}
M1 = {s ∈ B1(0)|0 ∈ A(s)},
where B1(0) is an open disc with s ∈ B1(0) ⇔ |s| < 1. The following will hold once Property 22 is proven:
|s| > 2−14 ⇒ |s2| >
√ 2
2 ⇒ s2 ∈ M ⇒ s ∈ M1.
The last of these implications will be proven with Proposition 22. We have:
A(s) = K + sK + s2K + . . . and by splitting the even and odd terms, we get:
A(s) = (K + s2K + s4K + . . .) + (sK + s3K + s5K + . . .)
= A(s2) + sA(s2).
Suppose A(s2) is connected and 0 /∈ A(s) = A(s2) + sA(s2). (This is a proof by contradiction).
If 0 /∈ A(s2) + sA(s2), also 0 /∈ A(s2).
Because 0 ∈ A(s2) implies that the element 0 + s · 0 ∈ A(s), so 0 ∈ A(s).
It follows that A(s2) ∩ sA(s2) = ∅.
This is because 0 /∈ A(s2), 0 /∈ A(s2) ∩ sA(s2) and if α ∈ A(s2) ∩ sA(s2), α ∈ A(s2) and α ∈ sA(s2). Then also −α ∈ A(s2) (change the sign of every an in a series) and so −α + α ∈ A(s).
It holds that sA(s2) ⊂ B, where B is a connected set in C − A(s2). The only possibility is that the set sA(s2) is in between 0 and A(s2).
Otherwise, for some α ∈ A(s2) and sα ∈ sA(s2), |sα| would be bigger than
|α|, so |s| > 1, but |s| < 1.
In the same way, it holds that the set s2A(s2) is in between 0 and sA(s2).
This would mean that
s2A(s2) ∩ A(s2) = ∅. (3)
Let L = s2A(s2), then A(s2) = (L + 1) ∪ (L − 1). Lemma 23 will lead to a contradiction to equation (3).
Lemma 23. "Wedge Lemma" Let L be a compact connected set. Suppose that (L + 1) ∩ (L − 1) 6= ∅. Then L ∩ ((L + 1) ∪ (L − 1)) 6= ∅.
Proof of Lemma 23 Suppose L is arc-connected. This means that L is path-connected: any two points a, b ∈ L can be connected with a continuous function f : [0, 1] → L, f (0) = a, f (1) = b, and that these functions are homeomorphisms.
Take
a = minl∈Ly(l) b = maxl∈Ly(l)
where y(l) is the y-value of an element of L in the complex plane.
Then B = {p = x + iy|a ≤ y ≤ b} is the smallest horizontal strip contain- ing L. It follows that B contains L, L + 1 and L − 1.
Let U and V be points where y(U ) = a and y(V ) = b. It follows that U − 1 and V + 1 are in (L + 1) ∪ (L − 1), which is arc-connected. ((L + 1) ∩ (L − 1) 6= ∅ and L is arc-connected).
The arc γ connects U − 1 and V + 1. Another arc γ0 connects U and V in L. These arcs necessarily intersect, as can be seen in figure 9.
This implies that at least one point, the intersection of γ and γ0 is in both (L + 1) ∪ (L − 1) and L. Thus L ∩ (L + 1) ∪ (L − 1) 6= ∅. (In the general case, where L is not necessarily arc-connected, we can just take an -neighborhood of L, which is arc-connected and let this go to zero.) This proves Lemma 23.
This prove used some concepts that were taught in the course analysis.
By Lemma 23, s2A(s2) ∩ A(s2) 6= ∅. Therefore, if A(s2) is connected, A(s) contains the element 0. Proposition 22 is now proven, because if it were not true, equation 3 was implied, and equation 3 has now been contradicted.
Figure 9: γ and γ0 intersect
Consequently, if s2 ∈ M , then s ∈ M1, so
|s| > 2−14 ⇒ s ∈ M1. (4) Recall that s ∈ M1 means that s ∈ B1(0) and 0 ∈ A(s).
So far it is known that for 2−14 < |s| < 1, there exists a sequence A(s) =
∞
X
i=0
aisi
with ai ∈ K = {−1, 1} and A(s) = 0. To prove Theorem 20, we need to show that the set M1 is equal to D ∩ B1(0). The sequences that give the elements of M1 as a solution should be "close to" Littlewood polynomials.
According to Lemma 15 in section 4, if the first n0−1 coefficients of a sequence A(s) =
∞
P
i=0
aisiand a Littlewood polynomial l(z) =
n
P
j=0
ajzjare the same, then there exists a zero of l(z) in an -neighborhood of each zero of A(s).
Take n0− 1 to be the degree n of the polynomial. Let A(s0) = 0, then for every , we can find a Littlewood polynomial of large enough degree n, such that a zero of this Littlewood polynomial, α has α ∈ B(s0). Therefore D ∩ B1(0) is dense in M1. So D ∩ B1(0) is equal to M1.
Proof of Theorem 20: According to Proposition 21, all 12√
2 < |s| < 1 have A(s) is connected. This means 12√
2 < |s| < 1 ⇒ s ∈ M . According to Proposition 22, A(s2) is connected implies that A(s) contains the element 0. This means that 2−14 < |s| < 1 ⇒ s ∈ M1. It has also been shown that D ∩ B1(0) is equal to M1.
It follows that n
2−14 < |z| < 1 o
⊂ D. According to Theorem 4, this implies that
n
2−14 < |z| < 1 o
∪n
1 < |z| < 214 o
⊂ D. Because the closure D is closed, it follows that
n
2−14 ≤ |z| ≤ 214o
⊂ D.
This proves Theorem 20.
6 Properties to be studied
In this section some other aspects of Littlewood polynomials will be mentioned.
In [10] polynomials with coefficients from a finite set are considered. One of these sets is {−1, 1}, which gives Littlewood polynomials. Some of the same results as in this thesis are considered, and a question is asked: What is the minimum number of zeros of modulus 1 of a reciprocal Littlewood polynomial?
This might be an interesting question. If it is known for a Littlewood polynomial with a certain degree, how many zeros it has of modulus 1, these can be controlled. Also, the multiplicity of certain zeros can be interesting to study.
The authors of [7], John Baez, Sam Derbyshire and Dan Christensen, have found the following remarkable property: Around a particular point z0, the set of roots of Littlewood polynomials near z0 looks like a scaled version of the iterated system corresponding to evaluating Littlewood polynomials near z0. The iterated system corresponds to the set A(z0), as in section 5.
The following pictures show the resemblance. With the correct scaling, the pictures show a better resemblance, as Baez et al. show.
Figure 10: Plot D around 0.5 + 0.5i Figure 11: A(0.5 + 0.5i)
In section 3 we saw that the only three real positive zeros in sets Q(√ d) with d ∈ Q>0 are
(
−1 −√ 5
2 , 1,1 +√ 5 2
) .
However, it seems (in figure 1 of section 1) as though the intersection D ∩ R>0is the interval [12, 2]. If this would be true, all rational numbers would be limit points of sequences in D. This is why we tried to find a sequence of Littlewood polynomials that would have roots closer and closer to 32. (Just to try for a rational number.) We did not succeed in this and others have tried as well without a result. The following polynomial of degree 55, had a zero very close to 1.5, however we were not able to make a sequence out of it.
−x55+ x54− x53+ x52+ x51+ x50+ x49+ x48− x47+ x46+ x45− x44+ x43 +x42− x41+ x40+ x39− x38+ x37+ x36− x35+ x34+ x33− x32− x31− x30 +x29− x28− x27− x26+ x25− x24− x23− x22− x21+ x20− x19− x18− x17
−x16+ x15− x14− x13− x12− x11− x10− x9− x8− x7− x6+ x5− x4− x3
−x2− x − 1
If we would be able to make a sequence as in the proof that 2 is a limit point, Theorem 11 in section 2:
`(z) = zn−1 − xn
1 − x = xn+1− xn+ 1 − xn
x − 1 = xn(x − 2) + 1 x − 1 for every rational number, then it would follow that
D ∩ R>0 = [1 2, 2].
This can be studied further. In particular, is it true, as indicated in figure 1 of section 1, that
D ∩ R>0 ?
= [1 2, 2].
7 Conclusion
The zeros of Littlewood polynomials make some beautiful pictures and they have a lot of properties which can be studied. In this thesis some properties of the set of zeros of Littlewood polynomials (D) were shown. Two known results were shown: the fact that the closure of the set of zeros of Littlewood polynomials, D is connected and that this set in fact contains all the complex numbers z with 2−14 ≤ |z| ≤ 214.
Also, the sets Q(√
d) for d ∈ Q were considered and in particular the intersection of these sets with D. The following result was found:
D ∩
∪
d∈QQ(
√ d)
=
±1, ±i, ±1 2±1
2
√−3, ±1 2±1
2
√ 5
. According to this result, only three elements of sets Q(√
d) with d ∈ Q are in D ∩ R>0.
The proofs that are given in this thesis make use of group theory, measure theory, analysis, algebraic structures, complex analysis, nearly all courses of the bachelor Mathematics.
For further research about Littlewood polynomials, the aspects mentioned in section 6 might be interesting. Also, it could be interesting to look at properties of polynomials with coefficients from the set {−1, 0, 1}. The zeros of these polynomials with degree up to 30 are plotted in the figure below.
I would like to thank Professor J. Top for supervising, and working on this project with me. Also I would like to thank mr. E.R. Duarte for writing the code that made figures 1-4, 6-8, 10 and 11 possible.
References
[1] http://www.pyth.eu/nummer.php?id=399
[2] https://en.wikipedia.org/wiki/Littlewood_polynomial [3] http://www.ma.man.ac.uk/ nikita/multinacci.html [4] Thierry Bousch, Paires de similitudes, January 1988.
[5] E. Bakker, Inversoren, http://irs.ub.rug.nl/dbi/4c860c8056f2e, 17-18, 2010.
[6] B. van Geemen, H.W. Lenstra, F. Oort, with additions from J.Top, Algebraïsche structuren, February 2014.
[7] John Baez, Dan Christensen and Sam Derbyshire, The beauty of roots, December 2011.
[8] E.B. Saff and A.D. Snider, Fundamentals of Complex Analysis with Ap- plications to Engineering, Science and Mathematics, Pearson education, 2003.
[9] Wilson A. Sutherland, Introduction to Metric & Topological Spaces, second edition, Oxford mathematics, 2009.
[10] J. Baradaran, M. Taghavi, communicated by S. Kanas, Polynomials with coefficients from a finite set, Mathematica Slovaca 64, 1397-1408, 2014.
[11] J. Top, Groepentheorie, 2013.
